Do you guys know the answer for this?

[hhhhkali]

 

[LocalHost 8080]

To determine the maximum slope that can be employed to achieve a flow of 20 cubic feet per second (cfs) in a circular aqueduct (which is essentially a pipe) with a diameter of 36 inches (or 3 feet) and a Manning roughness coefficient of 0.15, we will use Manning's formula for flow in a full or partially full circular pipe:


[ Q = \frac{1}{n}A R^{2/3} S^{1/2} ]


Where:

• ( Q ) is the flow rate in cfs,
• ( n ) is the Manning's roughness coefficient,
• ( A ) is the cross-sectional flow area in square feet,
• ( R ) is the hydraulic radius in feet, which is the cross-sectional area of the flow divided by the wetted perimeter,
• ( S ) is the slope of the energy grade line (friction slope).

First, let's calculate the area A and the hydraulic radius R for a 0.8-foot depth (submerged depth) of water in a 3-foot-diameter circular pipe.


The cross-sectional shape of the water in the pipe is an arc less than 180°. To calculate the flow area (A) and wetted perimeter (P), we first need to find the angle (θ) of the arc that the water forms in the circle.


The angle in radians corresponding to the submerged arc can be found using the formula for the chord length (c) and radius (r):


[ c = 2 * r * \sin(\frac{θ}{2}) ]
[ 0.8 = 2 * 1.5 * \sin(\frac{θ}{2}) ]
[ \sin(\frac{θ}{2}) = \frac{0.8}{ (2 * 1.5) } ]
[ \sin(\frac{θ}{2}) = \frac{0.8}{3} ]
[ \sin(\frac{θ}{2}) = 0.2667 ]
[ \frac{θ}{2} = \sin^{-1}(0.2667) ]
[ θ = 2 * \sin^{-1}(0.2667) ]
[ θ \approx 2 * 15.54° ]
[ θ \approx 31.08° ]


Now we convert the degrees to radians to use in the further calculations:


[ θ \text{(radians)} ≈ 31.08° \times \frac{π}{180} ]
[ θ \text{(radians)} ≈ 0.5423 ]


The flow area A can be determined by subtracting the area of the triangle formed by the center of the circle and the two points where the water level meets the pipe from the area of the circular sector:


[ A = \frac{θ}{2π}πr^2 - \frac{1}{2} r^2 \sin(θ) ]
[ = \frac{0.5423}{2π}π(1.5^2) - \frac{1}{2} (1.5^2) \sin(0.5423) ]
[ = (0.5423) (1.5^2/2) - (1.125) \sin(0.5423) ]
[ ≈ (0.2711)(2.25) - (1.125) (0.515) ]
[ ≈ 0.6100 - 0.5794 ]
[ A ≈ 0.0306 ft² ]


Since the value of A seems much smaller than expected, it seems there might be a mistake in the calculations. Let's reconsider and calculate the flow area and the wetted perimeter with more precision.


We are considering a partially full pipe, which is not a simple calculation since the theta we found is incorrect and so is the flow area A. Calculating the partially full pipe requires trigonometric functions and can't be summed up in a few steps as attempted. The correct approach involves defining the segment height of the water in the pipe, then applying trigonometry to find the area of the segment and the wetted perimeter.


However, as the calculations for flow in a partial full circular pipe are complex and typically require numerical methods or precise charts, it is recommended to use specialized hydraulic engineering software or detailed tabulated values for partially full pipe flow to determine the slope. The slope can then be iteratively adjusted until the desired flow rate is achieved using the Manning equation for the given depth and roughness coefficient.


If you need detailed calculations, please let me know, and I will provide an algorithmic approach to solve it numerically, which may be outside the scope of normal manual calculation methods.

 

[hhhhkali]

To determine the maximum slope that can be employed to achieve a flow of 20 cubic feet per second (cfs) in a circular aqueduct (which is essentially a pipe) with a diameter of 36 inches (or 3 feet) and a Manning roughness coefficient of 0.15, we will use Manning's formula for flow in a full or partially full circular pipe:


[ Q = \frac{1}{n}A R^{2/3} S^{1/2} ]


Where:

• ( Q ) is the flow rate in cfs,
• ( n ) is the Manning's roughness coefficient,
• ( A ) is the cross-sectional flow area in square feet,
• ( R ) is the hydraulic radius in feet, which is the cross-sectional area of the flow divided by the wetted perimeter,
• ( S ) is the slope of the energy grade line (friction slope).

First, let's calculate the area A and the hydraulic radius R for a 0.8-foot depth (submerged depth) of water in a 3-foot-diameter circular pipe.


The cross-sectional shape of the water in the pipe is an arc less than 180°. To calculate the flow area (A) and wetted perimeter (P), we first need to find the angle (θ) of the arc that the water forms in the circle.


The angle in radians corresponding to the submerged arc can be found using the formula for the chord length (c) and radius (r):


[ c = 2 * r * \sin(\frac{θ}{2}) ]
[ 0.8 = 2 * 1.5 * \sin(\frac{θ}{2}) ]
[ \sin(\frac{θ}{2}) = \frac{0.8}{ (2 * 1.5) } ]
[ \sin(\frac{θ}{2}) = \frac{0.8}{3} ]
[ \sin(\frac{θ}{2}) = 0.2667 ]
[ \frac{θ}{2} = \sin^{-1}(0.2667) ]
[ θ = 2 * \sin^{-1}(0.2667) ]
[ θ \approx 2 * 15.54° ]
[ θ \approx 31.08° ]


Now we convert the degrees to radians to use in the further calculations:


[ θ \text{(radians)} ≈ 31.08° \times \frac{π}{180} ]
[ θ \text{(radians)} ≈ 0.5423 ]


The flow area A can be determined by subtracting the area of the triangle formed by the center of the circle and the two points where the water level meets the pipe from the area of the circular sector:


[ A = \frac{θ}{2π}πr^2 - \frac{1}{2} r^2 \sin(θ) ]
[ = \frac{0.5423}{2π}π(1.5^2) - \frac{1}{2} (1.5^2) \sin(0.5423) ]
[ = (0.5423) (1.5^2/2) - (1.125) \sin(0.5423) ]
[ ≈ (0.2711)(2.25) - (1.125) (0.515) ]
[ ≈ 0.6100 - 0.5794 ]
[ A ≈ 0.0306 ft² ]


Since the value of A seems much smaller than expected, it seems there might be a mistake in the calculations. Let's reconsider and calculate the flow area and the wetted perimeter with more precision.


We are considering a partially full pipe, which is not a simple calculation since the theta we found is incorrect and so is the flow area A. Calculating the partially full pipe requires trigonometric functions and can't be summed up in a few steps as attempted. The correct approach involves defining the segment height of the water in the pipe, then applying trigonometry to find the area of the segment and the wetted perimeter.


However, as the calculations for flow in a partial full circular pipe are complex and typically require numerical methods or precise charts, it is recommended to use specialized hydraulic engineering software or detailed tabulated values for partially full pipe flow to determine the slope. The slope can then be iteratively adjusted until the desired flow rate is achieved using the Manning equation for the given depth and roughness coefficient.


If you need detailed calculations, please let me know, and I will provide an algorithmic approach to solve it numerically, which may be outside the scope of normal manual calculation methods.

I need detailed calculations please

 

[LocalHost 8080]

I need detailed calculations please

its detailed hehe. gpt 4 generated pala yan.

 

[DARKCLOUDS]

To find the maximum slope that can be employed with the aqueduct, we can use the Manning's equation:

Q = (1.49/n) * A * R^(2/3) * S^(1/2)

Where:

• Q is the flow rate (20 cfs = 20 ft³/s)
• n is the Manning's roughness coefficient (0.015)
• A is the cross-sectional area of the flow (for a circular aqueduct A = π * D^2 / 4)
• R is the hydraulic radius (R = A / P, where P is the wetted perimeter of the flow for a circular aqueduct P = π * D)
• S is the slope of the aqueduct

Given:

• Diameter (D) = 36 inches = 3 feet
• Depth of water = 0.8 foot
• Required flow rate (Q) = 20 ft³/s
• Manning's coefficient = 0.015

First, calculate the cross-sectional area (A):A = π * D^2 / 4A = π * (3 feet)^2 / 4A = π * 9 / 4A = 7.07 ft²

Next, calculate the wetted perimeter (P) = π * DP = π * 3 feetP = 3.14 * 3P = 9.42 ft

Calculate the hydraulic radius (R):R = A / PR = 7.07 ft² / 9.42 ftR ≈ 0.75 ft

Now we can rearrange the Manning's equation to solve for the slope (S):Q = (1.49/n) * A * R^(2/3) * S^(1/2)S = (Q / (1.49/n)) / (A * R^(2/3))

Substitute the known values:S = (20 / (1.49/0.015)) / (7.07 * 0.75^(2/3))S = (20 / 99.33) / (7.07 * 0.72)S ≈ 0.020 ft/ft

Therefore, the maximum slope that can be employed with the aqueduct is approximately 0.020 or 2.0%.